During electrolysis of water the volume of `O_(2)` liberated is `2.24 dm^(3)`. The volume of hydrogen liberated, under same conditions will be

Options

A. `2.24 dm^(3)`
B. `1.12 dm^(3)`
C. `4.48 dm^(3)`
D. `0.56 dm^(3)`

Answer

C

Explanation

`H_(2)O to H_(2) + (1)/(2)O_(2)`
During electrolysis, volumes of `O_(2)` and `H_(2)` liberated are in the ratio of 1 : 2 Hence, 3 volume of `H_(2)` liberated = `2 xx 2.24 dm^(3)= 4.48 dm^(3)`.